Two wires `A` and `B` of same material and same mass have radius `2r` and `r`. If resistance of wire `A` is `34 Omega`, then resistance of `B` will be

Options

A. `544 Omega`
B. `272 Omega`
C. `68 Omega`
D. `17 Omega`

Answer

A

Explanation

`R = (rho L)/A rArr R prop l/A`
Mass(m) = Volume × density = A l d
Since, there is same material of wire so ρ and d is-
same for both
`therefore Al`= constant
`l_(1)/l_(2) = A_(2)/A_(1)`
`R_(1)/R_(2) = l_(1)/l_(2) xx A_(2)/A_(1) =(A_(2)/A_(1))^(2) = (r_(2)/r_(1))^(4) (therefore I prop 1/A)`
`rArr 34/R_(2) = (r/(2r))^(4) rArr R_(2) = 544 Omega`